Problem 2136

Solution:

If $\sqrt[3]{3}$ had only the digit 2 between the 1000000th and the 3141592th digit after the decimal point, then with $a=\left[10^{1000000} \sqrt[3]{3}\right]<10^{1000001}$ we would have $\left|\sqrt[3]{3}-(a+2 / 9) 10^{-1000000}\right|<10^{-3141592}$, or equivalently

$$\left|\sqrt[3]{3\left(9 \cdot 10^{1000000}\right)^{3}}-(9 a+2)\right|<9 \cdot 10^{-2141592}$$

Now certainly $(9 a+2)^{3} \neq 3\left(9 \cdot 10^{1000000}\right)^{3}$, since the right-hand side contains the prime factor 3 with a multiplicity not divisible by 3. In general, for $m, n \in \mathbb{N}$ with $m \neq n^{3}$ we have:

$$|\sqrt[3]{m}-n| \geq \frac{\left|m-n^{3}\right|}{\sqrt[3]{m^{2}}+\sqrt[3]{m} n+n^{2}} \geq \frac{1}{3 \cdot(\max (\sqrt[3]{m}, n))^{2}}$$

From this, however, it follows - in contradiction to (1) -:

$$\left|\sqrt[3]{3\left(9 \cdot 10^{1000000}\right)^{3}}-(9 a+2)\right|>\frac{1}{3\left(10^{1000001}\right)^{2}}>10^{-2000003}$$